A chemical equation is a symbolic representation of all of
the substances involved in a chemical reaction. We use the
chemical formulas of substance to represent each chemical specie
involved in the reaction. We also use the notation (g), (l), (s),
or (aq) following the chemical formula to identify the phase of
the substances in the equation.

As an example of a chemical reaction we watched the reaction
between iron, Fe(s), and sulfur, S8(s), on video.
There are some additional
reactions on our Web site that you can review. We noted the
physical properties of both iron and sulfur before the reaction
and the physical property of the product of the reaction. The
magnetic property of iron, present when iron was in its elemental
form, was absent in the product. The yellow color of the sulfur,
present in the elemental form as a reactant was absent in the
product. Yet both iron and sulfur are in the product. Since the
properties changed the product is a new substance.

How do we tell if a chemical reaction occurs? The best
evidence is a difference in the physical and chemical properties
of the reactants and the products. This can be obvious as in the
case of the reaction between iron and sulfur. Other
characteristics to watch for when mixing two or more substances;

heat and light (flames)

explosion

color change

evolution of a gas

formation of a solid (precipitate)

The form of a chemical equation involves writing the formulas
of the reactants (the substances that are mixed together) on the
left, using '+' when more than one substance is involved and the
formula(s) of the product(s) on the right. The reactants and the
products are separated by an arrow '--->'. Sometimes
additional information about the reaction is placed above or
below the arrow which separates the reactants and products. Such
information include;

heat (using a delta symbol)

temperature (at which the reaction is run at)

pressure

time (length of time the reaction is allowed to
proceed)

Well, how about writing the equation for the reaction between
iron and sulfur? To do that we have to know the formula for both
elements and their phases;

Fe(s) + S8(s) --heat-->

(Note: I can not use a delta symbol on the WEB
easily). Now we have to know what the products are. This is a
little more difficult but in this case is it easy. We are
reacting a metal with a nonmetal so we know the product is an
ionic compound. Since there are only two elements reacting the
product has to be binary. So we need the formula for a binary
ionic compound. You know how to do that. Metals form cations and
the nonmetal forms an anion and we need to use the principle of
electroneutrality to balance the charges. Our only other problem
is that iron is a transition metal and as such it can have
several different charges. For our purposes we limit the
possibilities to 2+ or 3+. If we use 2+ for iron and we know
sulfur is always 2-, the formula of the product would have to be,
FeS(s). We know it is a solid because we saw on the video it was
solid, but also ALL ionic compounds are solids!

So the reaction is,

Fe(s) + S8(s) --heat--> FeS(s)

Another reaction (side reaction) which occurred
as the above reaction occurred was between sulfur and oxygen in
the air. This side reaction was evident because of the bluish
flame that appeared as the sulfur melted. At one point in the
video as we looked the crucible, containing the reaction, from
the side the blue flame was erupting from the crucible. The
reaction was surpressed when the hand placed the crucible lid
onto the crucible. But what was the reaction?

S8(s) + O2(g)
--heat-->

This is a harder question because we are
combining two nonmetals. The product of this reaction we know
will be a binary covalent compound. But we do not know what the
formula of the product might be because we have not discussed any
rules for determining formulas of covalent compounds. So what do
we do? Well, unfortunately, we have to know some formulas of
binary compounds containing sulfur and oxygen. You know sulfate,
SO42-, and sulfite, SO32-
are either of these likely? No! And the reason is the charged
nature of those species. We will not form a charged substance in
this particular case. What else you ask? How about SO? Is that
possible? Well, know it is not. If we try to find such a compound
in the Merck Index or the CRC we will not. So how about SO2?
In this case you would be right on! That is the correct formula.
Sulfur dioxide is an evil smelling, colorless, gas. So the
reaction is;

S8(s) + O2(g)
--heat--> SO2(g)

These two reactions we have discussed are called
formation reactions. (Our textbook also refers to these reactions
as synthesis or combination reactions.)

Formation reactions are characterized by the
fact that the reactant are elements in their standard state and
the product is a compound containing the reactant elements
combined in some way. We have seen two formation reactions so
far.

Fe(s) + S8(s)
--heat--> FeS(s)

S8(s) + O2(g)
--heat--> SO2(g)

Formation reactions involving metals and
nonmetals are the more straightforward. Writing the formula of
the product is guided by the knowledge that the metal becomes
positively charged and the nonmetal becomes negatively charged.
We use the principal of electroneutrality (balance the charges)
to write the formula of the product. When the elements are both
nonmetals predicting the products is a little more difficult.
Besides the one example of this type above, others we have
discussed include;

Na(s) + Cl2(g) ---->
NaCl(s)

N2(s) + O2(g)
--heat--> 2NO2(g)

C(s) + O2(g) ----> CO2(g)

P4(s) + 5O2(g)
--heat--> P4O10(s)

Several of these formation reaction we have
discussed are not balanced. So let's take a moment and talk about
balancing chemical equations.

How about balancing these reactions? Balancing
equations is the process of equating the elements on both sides
of the reaction arrow. A balanced equation has equal numbers of
each element on both sides of the equation. Balancing consists of
introducing coefficients which proceed the formula in the
equation. Subscripts are NEVER changed when balancing equations.

For many reactions a method of trail and error
is used. By practicing you're skills will sharpen. As you balance
more and more equations certain 'rules' begin to emerge which can
be useful. One rule which is very useful is to balance elemental
forms last when balancing equations. This rule derives from the
fact that changing the coefficient preceding an element only
effects that element. Changing a coefficient preceding a compound
changes all the elements in the compound and may effect another
substance in the equation.

So shall we balance a few of the equations we've
covered so far?

Let's begin with;

Fe(s) + S8(s) --heat--> FeS(s)

In this equation if we begin with the element
iron, we see there one Fe atom on the left and one Fe atom on the
right. So iron is balanced. Sulfur has eight atoms on the left
and only one on the right. Sulfur is not balanced. To balance the
sulfur atoms we need to replace the coefficient preceeding FeS
with the number 8.

Fe(s) + S8(s) --heat--> 8FeS(s)

Now the sulfur atoms are balanced on both sides,
but by adding the coefficient we've unbalanced the iron atoms. To
fix that we need to place a coefficient of 8 before the elemetn
Fe on the left side of the equation.

8Fe(s) + S8(s)
--heat--> 8FeS(s)

Now the equation is balanced.

Balance the following equations;

We've used the reaction between iron and sulfur
and between sulfur and oxygen as examples of formation reactions.
There was another reaction which occurred when the iron and
sulfur mixture was heated. That reaction occurred when we lit the
Bunsen burner to heat the mixture of iron and sulfur. The
reaction that was used to heat the mixture was a combustion
reaction between methane (CH4) and oxygen (O2).

CH4(g) + O2(g)
--heat-->

Methane is an example of a hydrocarbon. It is
not the only example of a hydrocarbon. But more about that
shortly. The products of this combustion reaction are carbon
dioxide gas and water vapor (liquid).

CH4(g) + O2(g)
--heat--> CO2(g) + H2O(g)

Other hydrocarbons you are familiar with
include;

Name

Formula

methane

CH4(g)

ethane

C2H6(g)

propane

C3H8(g)

butane

C4H10(g)

pentane

C5H12(l)

hexane

C6H14(l)

heptane

C7H16(l)

octane

C8H18(l)

nonane

C9H20(l)

decane

C10H22(l)

You should know the name and formula for the
first ten hydrocarbons (memorize).

Methane is a colorless, odorless gas which is
used as a fuel in most gas stoves to heat/cook food. Gas
companies add a compound which has an odor to help detect gas
leaks should they arise. The compound I mentioned in class that
was added to natural gas was CH3SH. This is an error,
the compound that is added to natural gas is call t-butyl
mercaptan and has the formula C4H9SH. So
lets write the combustion reaction of this compound.

Hey we get more practice balancing equations!

Balance the following equations;

Double Displacement/Replacement Reactions

All ionic compounds are solids in their standard state (25 degrees
Celsius and 1 atmosphere). Ionic compounds are composed of ions; elements or
molecules which are positively charged are called cations and elements or molecules
which are negatively charged are called anions. A simple model to be used to
imagine what ions (cations or anions) look like at the atomic level, is a sphere.
For simple monoatomic ions and most polyatomic ions we can think of as charged
spheres of different sizes.

In the state the spheres (ions) are arranged so that any particular
ion is surrounded by the oppositely charged ion. What ever pattern which arises
depends on several factors which we will not be discussing at this point in
our class. Below is a picture of a model of a simple ionic compound. Notice
the spheres (ions) of different sizes and how each sphere (ion) is surrounded
by the other spheres (ion).

What happens when an ionic compound dissolves in water? Below
is a figure of a small crystal of sodium chloride surrounded by water molecules.
We all know when sodium chloride is added to water the crystals of sodium chloride
disappear, that is, they dissolve.

What happens when the sodium chloride dissolves? Below is a picture
showing a chloride ion (larger sphere) and a sodium ion (smaller sphere) dissolved
in water.

Notice how water surrounds (hydrates) each ion. For the negatively
charged chloride ion the hydrogen's of each water molecule orient themselves
towards the chloride ion. For the positively charged sodium ion the oxygen's
of each water molecule orient themselves towards the sodium ion.

The specific example of what happens when sodium chloride is
added to water can be used for the general case of any ionic compound.

Remember we talked about the conductivity of solutions containing
dissolved ionic compounds. You will be testing the conductivity of soluble ionic
compounds in laboratory later this semester.

To begin we looked at the conductivity of a sample of deionized water
(water with little or no ions) which could be viewed as pure water. Looking
at the image on the left the light bulb is not glowing. This can be explained
in terms of the nature of the species in the sample of water. That the
light does not glow suggests there are no ions in the sample. This is
an important measurement to establish subsequent measurements in deionized
water. Any observed conductivity will due to the presence of the solute.

Next we measured the conductivity of an aqueous solution of sodium chloride.
Here the light bulb glows brightly indicating the presence of ions in
the solution. The ability of an aqueous solution of sodium chloride to
conduct electricity is characteristic of a strong electrolyte. How do
we express this observation using a chemical equation?

NaCl(s) ---H2O--> Na+(aq)
+ Cl-(aq)

Soluble ionic compounds are also strong electrolytes, i.e., behave the
same way as sodium chloride...forming ions when dissolved in water. How
do we know whether an ionic compound dissolves in water? That is determined
by experiment. A Solubility Table summarizes the experimental observations
of the solubility behavior of a large group of ionic compounds. We'll
discuss a Solubility Table shortly.

When we tested the conductivity of an aqueous solution of sucrose the
light bulb did not glow. This type of behavior is characteristic of a
nonelectrolyte. No ions in this solution, yet the sucrose dissolved. How
do we write a chemical equation in this case?

C12H22O11(s) ---H2O-->
C12H22O11(aq)

In general soluble covalent compounds do not dissociate into ions when
dissolved in water.

In laboratory you will investigated the solubility of a large collection of
ionic compounds. For those ionic compounds that dissolved in water you were
expected to write the formula of ions formed. In the post-laboratory you were
to write a chemical equation describing what happens when a soluble ionic compound
dissolves in water.

For example, sodium chloride dissolves in water. The chemical equation that
can be written to describe the solubility of sodium chloride is,

NaCl(s) ---H2O--> Na+(aq)
+ Cl-(aq)

We can write the equation to describe what happens when copper(II) chloride
dissolves in water,

CuCl2(s) ---H2O--> Cu2+(aq)
+ 2Cl-(aq)

You must be able to write the ions formed and the chemical equation which
describe the behavior of soluble ionic compounds.

The goal of Experiment #2 is to organize all the experimental observations
into something meaningful. This data can be organized into a Solubility Table.
The Solubility Table summarizes the solubility behavior of ionic compounds in
terms of anions and cations.

A Solubility Table summarizes the solubility behavior of a large group of
ionic substances.

How to interpret a Solubility Table?

We can use the Solubility Table to determine whether an ionic compound exist
as ions in aqueous solution (soluble) or as a solid (insoluble). Once we know
the compound we use the Solubility Table to determine its solubility.

For example, consider the following compounds; NaCl, BaSO4, NaC2H3O2,
and CaS. Determine the solubility in water for these ionic substances.

The solubility of each of these compounds can be determined by using the Solubility
Table.

NaCl (all chlorides are soluble except...) SOLUBLE;

BaSO4 (all sulfates are soluble except...) INSOLUBLE;

NaC2H3O2 (all sodium compounds are soluble)
SOLUBLE;

CaS (all sulfides are insoluble...) INSOLUBLE

For those compounds that are soluble write a chemical equation which describes
what happens when the solid is added to water.

NaCl(s) ---H2O--> Na+(aq) +
Cl-(aq)

NaC2H3O2(s) ---H2O--> Na+(aq)
+ C2H3O2-(aq)

We'll also use the information in a Solubility Table to help identify the
phase of ionic substance in a chemical equation. The chemical reaction types
where the Solubility Table is important are;

Double Replacement reactions

Neutralization reactions

Single Replacement reactions

Lets consider a double displacement reaction problem;

Write the formula and identify the phase for the product(s) and balance the
following reaction.

Na2SO4(aq) + CaCl2(aq) --->

Since this is a double replacement reaction we can write the formulas of the
products by exchanging the cations and anions. When you do this remember to
keep track of the individual charges on the cations and anions.

Na2SO4(aq) + CaCl2(aq) ---> CaSO4(?)
+ 2NaCl(?)

Now we'll use the Solubility Table to predict the phases of the products.
According to the table CaSO4 is INSOLUBLE and NaCl is SOLUBLE.

Na2SO4(aq) + CaCl2(aq) ---> CaSO4(s)
+ 2NaCl(aq)

Let's consider another example.

Write the formula and identify the phase for the product(s) and balance the
following reaction.

AgNO3(aq) + Na2CO3(aq) --->

Since this is a double replacement reaction we can write the formulas of the
products by exchanging the cations and anions. When you do this remember to
keep track of the individual charges on the cations and anions.

2AgNO3(aq) + Na2CO3(aq) --->Ag2CO3(?)
+ 2NaNO3(?)

Now we'll use the Solubility Table to predict the phases of the products.
According to the table Ag2CO3 is INSOLUBLE and NaNO3
is SOLUBLE.

To change the above equation to an ionic equation the aqueous ionic substances
must be written as ions;

The net ionic equation is;

2Ag+(aq) + CO32-(aq)
---> Ag2CO3(s)

Neutralization Reactions

A reaction between an acid and a base to produce a salt and water. (Remember
you will need to be able to write the ionic and net ionic equations for
these reactions.)

To begin with we need to be able to list some acids and bases that were are
familiar with...here are a few I think everyone should know...already knows
from our discussion in Chapter 6 on nomenclature.

Important Acids

Name

Formula

Sulfuric acid

H2SO4

Sulfurous acid

H2SO3

Nitric acid

HNO3

Nitrous acid

HNO2

Phosphoric acid

H3PO4

Phosphorus acid

H3PO3

Carbonic acid

H2CO3

Perchloric acid

HClO4

Acetic acid

HC2H3O2

Formula

Name

HF(aq)

Hydrofluoric acid

HCl(aq)

Hydrochloric acid

HBr(aq)

Hydrobromic acid

HI(aq)

Hydroiodic acid

H2S(aq)

Hydrosulfuric acid

HCN(aq)

Hydrocyanic acid

Important Bases (All of the Group IA and IIA hydroxides)

Name of Base

Formula of Base

Sodium hydroxide

NaOH

Potassium hydroxide

KOH

Barium hydroxide

Ba(OH)2

Ammonia

NH3

Calcium hydroxide

Ca(OH)2

Aluminum hydroxide

Al(OH)3

Let's consider a few neutralization reactions and how we write the equations.

Consider the reaction between hydrochloric acid and sodium hydroxide;

HCl(aq) + NaOH(aq) --->

To write the products we combine the anion of the acid with the cation of
the base and write the correct formula following the principle of electroneutrality.
The other product is water.

Molecular equation: HCl(aq) + NaOH(aq)
---> NaCl(aq) + H2O(l)

So the molecular form of the equation is shown above. To write the ionic equation
we must separate all aqueous species into their ions and leave any solid, liquid
or gaseous substance in its molecular form. So in this case HCl(aq),
NaOH(aq), and NaCl(aq) must be written as aqueous ions
and H2O(l) remains in its molecualr form.

Finally to write the net ionic equation we must cancel all species common
to both sides of the equation

net ionic equation: H+(aq) + OH-(aq)
---> H2O(l)

Consider the reaction between nitric acid and calcium hydroxide;

HNO3(aq) + Ca(OH)2(aq) --->

To write the products we combine the anion of the acid with the cation of
the base and write the correct formula following the principle of electroneutrality.
The other product is water. Be sure to balance this equation.

So the molecular form of the equation is shown above. To write the ionic equation
we must separate all aqueous species into their ions and leave any solid, liquid
or gaseous substance in its molecular form. So in this case HNO3(aq),
Ca(OH)2(aq), and Ca(NO3)2(aq) must be written
as aqueous ions and H2O(l) remains in its molecualr form.

Finally to write the net ionic equation we must cancel all species common
to both sides of the equation

net ionic equation: 2H+(aq) + 2OH-(aq)
---> 2H2O(l)

or

net ionic equation: H+(aq) + OH-(aq)
---> H2O(l)

Consider the reaction between hydrobromic acid and ammonia;

HBr(aq) + NH3(aq) --->

To write the products we combine the anion of the acid with the cation of
the base and write the correct formula following the principle of electroneutrality.
The other product is water. Be sure to balance this equation.

Molecular equation: HBr(aq) + NH3(aq)
---> NH4Br(aq)

Notice in this particular neutralization equation no water is formed. Since
there is no hydroxide ions we can not write water as a product. So when ammonia
is one of the reactants we do not include water as a product. Water is present
since the reaction occurs in aqueous solution, we just do not write it as a
product.

So the molecular form of the equation is shown above. To write the ionic equation
we must separate all aqueous species into their ions and leave any solid, liquid
or gaseous substance in its molecular form. So in this case HBr(aq)
and NH4Br(aq) must be written as aqueous ions and NH3(aq)
remains in its molecualr form.

ionic equation: H+(aq) + Br-(aq)
+ NH3(aq) ---> NH4+(aq) + Br-(aq)

Finally to write the net ionic equation we must cancel all species common
to both sides of the equation

ionic equation: H+(aq) + Br-(aq)
+ NH3(aq) ---> NH4+(aq) + Br-(aq)

net ionic equation: H+(aq) + NH3(aq)
---> NH4+(aq)

Consider ONE more reaction, between sulfuric acid and barium hydroxide;

H2SO4(aq) + Ba(OH)2(aq) --->

To write the products we combine the anion of the acid with the cation of
the base and write the correct formula following the principle of electroneutrality.
The other product is water. Be sure to balance this equation.

Molecular equation: H2SO4(aq) + Ba(OH)2(aq)
---> BaSO4(s) + 2H2O(l)

So the molecular form of the equation is shown above. To write the ionic equation
we must separate all aqueous species into their ions and leave any solid, liquid
or gaseous substance in its molecular form. So in this case H2SO4(aq)
and Ba(OH)2(aq) must be written as aqueous ions and BaSO4(s)
and 2H2O(l) remains in their molecualr form.

Finally to write the net ionic equation we must cancel all species common
to both sides of the equation. But there are no species common to both sides
of the equation!

So the net ionic equation and the ionic equaton are the same.

Looking at the list of acids and bases at the top of the page
you can imagine ALL the possibilities. So practice a few on your own until you
get comfortable with writing neutralization equations.

Single replacement Reactions

A reaction in which an element in a reactant compound is replaced by a
second reacting element producing a new compound and an element. (Remember
you will need to be able to write the ionic and net ionic equations.)

2Li(s) + 2H2O(l) ---> 2LiOH(aq)
+ H2(g)

2Na(s) + 2H2O(l) ---> 2NaOH(aq)
+ H2(g)

2K(s) + 2H2O(l) ---> 2KOH(aq)
+ H2(g)

Mg(s) + 2HCl(aq) ---> MgCl2(aq)
+ H2(g)

We looked at the reactions of lithium with water, sodium with water, potassium
with water and barium with water. All of these are examples of single replacement
reactions and following the equation as show about.

We also looked at the reaction of magnesium with hydrochloric acid. In general
only the most reactive metal (Group IA) react with water. Group IIA element
are not as reactive with water. To get magnesium to react with water requires
the water to be hot. Calcium and barium both react with water, but the reactions
are not as impressive at the Group IA elements. To get the Group IIA elements
to liberate hydrogen we need to react the metals with an acid like hydrochloric
acid. Zinc will also react with hydrochloric acid like magnesium.

Decomposition Reactions

The only example of a decomposition reaction I showed was that of potassium
chlorate decomposing when heated.